Thanks for the logic and style, I'm sure many have learned from it! Following the same logic described, I have the same solution written in C++ for anyone who wanted to see it in the language

Here it is:

long substrCount(int n, string s) {

long count = 0;
vector<std::pair<char, int>> frequencies;
int i=0, j=0;

for ( i = 0; i < n; i++)
{
    for( j = i+1; j < n; j++)
    {
        if (s[j] == s[i])
            continue;
        else
            break;
    }
    frequencies.push_back(std::make_pair(s[i],j-i));
    i = j-1;
}

for (i=0; i < frequencies.size(); i++)
    count += (frequencies[i].second+1) * frequencies[i].second / 2;

for (i=1; i < frequencies.size()-1; i++)
{
    if ( frequencies[i].second == 1 && frequencies[i-1].first == frequencies[i+1].first)
        count += min(frequencies[i-1].second, frequencies[i+1].second);
}

return count;

}